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PhysicsSHMCalculation of time periodEasy2 minPYQ_2020

PhysicsEasysingle choice

The time period of a particle executing S.H.M. is $8 s$ . At $t = 0$ it is at the mean position. The ratio of the distance covered by the particle in the $1^{s t}$ second to the $2^{n d}$ second is

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Answer:

Solution:

$T = 8 s$

$ω = \frac{2 π}{T} = \frac{π}{4}$

Equation of S.H.M. $y = A \sin \frac{π}{4} t$

$y (1) = A \sin (\frac{π}{4}) = \frac{A}{\sqrt{2}}$

$y (2) = A \sin (\frac{π}{2}) = A$

Distance travelled in $2^{n d}$ second $= y (2) - y (1)$

$= A - \frac{A}{\sqrt{2}}$

So the ration would be $= \frac{A - \frac{A}{\sqrt{2}}}{\frac{A}{\sqrt{2}}}$

$= \frac{\sqrt{q} - 1}{1} = \frac{(\sqrt{2} - 1) (\sqrt{2} + 1)}{\sqrt{2} + 1} = (\frac{1}{\sqrt{2} + 1})$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: PYQ_2020

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